## mathmath333 one year ago functions

1. mathmath333

\large \color{black}{\begin{align} &\normalsize \text{Find the maximim value of the function}\hspace{.33em}\\~\\ & \dfrac{1}{x^2-3x+2} \hspace{.33em}\\~\\ \end{align}}

2. SolomonZelman

interval given ?

3. SolomonZelman

with out interval this maximum value diverges to infinity.

4. mathmath333

this question means that i have to minimize \large \color{black}{\begin{align} x^2-3x+2 \hspace{.33em}\\~\\ \end{align}} right ?

5. SolomonZelman

should be, or els... i mean there is not maximum value here, if you are taking over $$(-\infty,~+\infty)$$

6. SolomonZelman

wait

7. SolomonZelman

1/that ? or, just that ?

8. mathmath333

lol its \large \color{black}{\begin{align} \dfrac{1}{x^2-3x+2} \hspace{.33em}\\~\\ \end{align}}

9. SolomonZelman

$$\large\color{black}{ \displaystyle f(x)=\frac{1}{x^2-3x+2} }$$ and you want to find absolute minimum, right?

10. SolomonZelman

well, I can tell you that the limit as x approaches $$\pm$$infinity is going to be 0.

11. SolomonZelman

we can plot, I am guessing it is something like |dw:1435010239497:dw|

12. mathmath333

i want to find the maximum value of this \large \color{black}{\begin{align} \dfrac{1}{x^2-3x+2} \hspace{.33em}\\~\\ \end{align}}

13. SolomonZelman

none

14. SolomonZelman

it is +infinity

15. mathmath333

but how ?

16. SolomonZelman
17. mathmath333

how i prove that with algebra

18. SolomonZelman

even the minimum doesn't exist.

19. SolomonZelman

with algebra, even with no calc ?

20. mathmath333

idk calculus

21. SolomonZelman

ic

22. SolomonZelman

lets c

23. SolomonZelman

$$\large\color{black}{ \displaystyle f(x)=\frac{1}{x^2-3x+2} }$$ it has vertical asymptotes, can you find them for me?

24. mathmath333

\large \color{black}{\begin{align} x=2,\ 1 \hspace{.33em}\\~\\ \end{align}}

25. SolomonZelman

yes

26. SolomonZelman

and, the best thing I can see here is to use the idea of a limit to see where range tends to the close we get to x=1 and x=2. I don't really know a purely algebraic proof for why this function has no max or min.

27. xapproachesinfinity

that has a max actually

28. SolomonZelman

yes?

29. SolomonZelman

https://www.desmos.com/calculator keep scrollin up

30. SolomonZelman

on a certain interval, no doubt (unless it has a domain gap)

31. xapproachesinfinity

i think it attains one max does not perhaps required any domain to get it or i think

32. xapproachesinfinity

your graph does not show up see here https://www.desmos.com/calculator

33. SolomonZelman

i don't think there is a max. that is due to an asymptote - two of them, there.

34. SolomonZelman

we can find the limit from the right and from the left as x approaches 1 and 2.

35. xapproachesinfinity

http://prntscr.com/7k4q1l here see that bottom part of the graph

36. xapproachesinfinity

with calculus we can prove it has a max value but @mathmath333 is looking for a different way

37. SolomonZelman

this graph goes further down and up if you scroll on desmos, but we can go ahead and take any limit that is closed from either side of any of the asymptotes you wish to choose.....

38. xapproachesinfinity

i mean here local max not abolute of course

39. mathmath333

how to find that local max

40. SolomonZelman

oh, but the question is asking for absolute doesn't it?

41. SolomonZelman

that is what i would think of course

42. xapproachesinfinity

first let's write that as 1/(x-1)(x-2)

43. xapproachesinfinity

44. mathmath333

yes the question is asking absolute, i just asked that of curiosity

45. SolomonZelman

this function is pretty weird. I wonder what would it model

46. SolomonZelman

i mean in real world

47. xapproachesinfinity

you want to then prove there is no max algebraically

48. mathmath333

yes ^

49. SolomonZelman

would that include using ideas from calc but not its techniques?

50. SolomonZelman

I am stubborn on the idea of a limit. A limit would simply show that as it approaches __ it can be infinitely large/small

51. mathmath333

ok i will try to bear with calculus

52. SolomonZelman

$$\large\color{black}{ \displaystyle \frac{1 }{x^2-3x+2} }$$ we need to do only 1 asymptote for this

53. SolomonZelman

$$\large\color{black}{ \displaystyle \frac{1 }{x^2-3x+2} }$$ lets plug in values that are less than, but close to 1

54. SolomonZelman

$$\large\color{black}{ \displaystyle f(0.9)=? }$$

55. SolomonZelman

56. SolomonZelman

f(0.99)=99

57. SolomonZelman

try plugging it yourself for close values to 1

58. SolomonZelman

these values are called approaching "from the left" because they are a buit less than 1

59. SolomonZelman

now, try to do f(1.5) f(1.1) f(1.01) f(1.001) and you will see where it goes "from the right" - for values that are greater than 1

60. SolomonZelman

i got to go

61. SolomonZelman

sorry

62. xapproachesinfinity

suppose there exist a max max(1/(x^2-3x+2)=c 1/x^2-3x+2<c x^2-3x+2>1/c=l x^2-3x+2-l>0

63. xapproachesinfinity

just some random stuff lol

64. xapproachesinfinity

coming back..

65. mathmath333

ok i need to go please comment if u found something

66. xapproachesinfinity

let's first consider the domain $$D_f: \{ x\in \mathbb{R}: x\ne1, x\ne 2 \}$$

67. xapproachesinfinity

that is (-oo, 1)u(1, 2)u(2,oo)

68. xapproachesinfinity

we look at the sign of y when x<1 if y does not change the sign in that interval then i must not have a max seems that 1/(x-2)(x-1)>0 for x<1 same reasoning for x>2 ===> y>0 so only btw 1 and 2 left

69. xapproachesinfinity

from the graph we knew for 1<x<2, y<-4 we need to show this